A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is :

Solution

Physics sir · 4 steps

A rolling sphere both moves forward and spins, so add both kinetic energies and you get 7/10mv².

1. Two kinds of motion. A body rolling without slipping has translational KE 1/2mv² and rotational KE 1/2Iω². Rolling without slipping means v = ω r.

   K = 1/2mv² + 1/2(2/5mr²)(v/r)² = 1/2mv² + 1/5mv² = 7/10mv²

2. Convert to SI. m = 500 g = 0.5 kg and v = 5.00 cm s⁻¹ = 0.05 m s⁻¹, so v² = 2.5 × 10⁻³ m² s⁻².
3. Put in numbers. K = 0.7 × 0.5 × 2.5 × 10⁻³ = 8.75 × 10⁻⁴ J.
4. So. The kinetic energy is 8.75 × 10⁻⁴ J, which is option (c).

The idea to keep. For rolling, K = 1/2mv²(1 + k²/r²). For a solid sphere k²/r² = 2/5, so K = 7/10mv². Remember this factor and the question takes one line.

Answer from the NTA final answer key. Solution by PrepWiser.